# Can the zeroes of a multivariate $p$-adic polynomial be bounded?

Multivariate real polynomials, as opposed to multivariate complex polynomials, can have bounded zero sets, i.e. $x^2+y^2-1$ in $\mathbb{R}^2$. This fails in $\mathbb{C}^n$ because $\mathbb{C}$ is algebraically complete and restricting to a generic line will produce a univariate polynomial which will always have roots. Are there any polynomials like this in $\mathbb{Q}_p^n$ with $n>1$?

$\mathbb{Q}_p$ isn't algebraically closed so that argument doesn't immediately go through, but it also isn't ordered like $\mathbb{R}$ so it isn't obvious how to construct an example of a polynomial with bounded zero set. I know from this answer that in $\mathbb{Q_p}$ $-1$ is the sum of 4 squares, which I think implies that $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2-1$ does not have a bounded set of zeroes in $\mathbb{Q}_p^5$, which is at odds with the same case in $\mathbb{R}^5$, but doesn't tell me anything about the general situation.

• can you explain what would be an unbounded subset of $\Bbb Q_p$ ? – mercio Apr 6 '16 at 22:29
• @mercio OP probably means solutions whose coordinates have unbounded $p$-adic norm. – RKD Apr 6 '16 at 23:03
• @Jake is right. I meant unbounded norm. – James Hanson Apr 7 '16 at 2:23
• oh I was thinking about $\Bbb Z_p$ so it didn't make sense lol. Alright. – mercio Apr 7 '16 at 6:55

I was able to find a relevant source here. There are several examples in that document but I was primarily interested in the 'circle' in $\mathbb{Q}_p^2$. For primes of the form $p=4k+1$ the 'circle' is unbounded and for primes of the form $p=4k+3$ the 'circle' is bounded (I haven't worked out the $p=2$ case yet).
In $\mathbb{Q}_p$ with $p=4k+1$ there is a square root of $-1$ which I'll call $i$. This allows for the factorization $x^2+y^2=(x+iy)(x-iy)$ which with a change of basis means that the 'circle' is equivalent to the polynomial $uv-1$, which clearly has arbitrarily large zeroes.
In $\mathbb{Q}_p$ with $p=4k+3$ assume that you have a solution $x,y$ of $x^2+y^2=1$ with $|x|_p>1$. Then by the strong triangle inequality $|y|_p=|x|_p>1$. This implies that if we take the least significant digits of $x$ and $y$ (call them $x_0$ and $y_0$) in their base $p$ expansions then it must be the case that $x_0^2+y_0^2=0$ in $\mathbb{F}_p$. But this is only possible if $x_0$ and $y_0$ are both $0$, which is a contradiction. The same argument applies for $y$ so we have that both the x and y coordinates of the zeroes of $x^2+y^2-1$ are bounded and the set is bounded in $\mathbb{Q}_p^2$.